Find the gradient of $f(x, y) = \sqrt{x + y}$ at $(3, 6)$. $\nabla f(3, 6) = ($
Answer: The gradient of a scalar field is all its partial derivatives put together into a vector. For a 2D scalar field, this looks like $\nabla f = (f_x, f_y)$. Let's find $f_x$ and $f_y$. $\begin{aligned} f_x &= \dfrac{\partial}{\partial x} \left[ \sqrt{x + y} \right] \\ \\ &= \dfrac{1}{2 \sqrt{x + y}} \\ \\ f_y &= \dfrac{\partial}{\partial y} \left[ \sqrt{x + y} \right] \\ \\ &= \dfrac{1}{2 \sqrt{x + y}} \end{aligned}$ Now we can evaluate the partial derivatives we found at the point $(3, 6)$. $\begin{aligned} f_x(3, 6) &= \dfrac{1}{2 \sqrt{x + y}} = \dfrac{1}{2\sqrt{9}} = \dfrac{1}{6} \\ \\ f_y(3, 6) &= \dfrac{1}{2 \sqrt{x + y}} = \dfrac{1}{2\sqrt{9}} = \dfrac{1}{6} \end{aligned}$ The gradient of $f$ at $(3, 6)$ is $\nabla f = \left( \dfrac{1}{6}, \dfrac{1}{6} \right)$.